Lecture 32 Manipulating Power Series

Text References: Course notes pp. 137-139 & Rogawski 10.6

32.1 Recap

Last time, we learned about power series and how to find their radius of convergence.

Exercise 32.1 Find the values of \(x\) for which \(\displaystyle \sum_{k=0}^\infty \frac{(x-1)^k}{\sqrt{k+1}}\) converges.

Solution. Using the Ratio test, we have

\[\begin{align*} \lim_{k\to\infty}\left | \frac{a_{k+1}}{a_k}\right| &= \lim_{k\to\infty}\left | \frac{\frac{(x-1)^{k+1}}{\sqrt{k+2}}}{\frac{(x-1)^k}{\sqrt{k+1}}}\right | \\ &= \lim_{k\to\infty} \left | \frac{(x-1)^{k+1}}{(x+1)^k}\frac{\sqrt{k+1}}{\sqrt{k+2}}\right | \\ &= \lim_{k\to\infty} |x-1| \sqrt{\frac{n+1}{n+2}}\\ &= |x-1| \end{align*}\]

Therefore, by the Ratio Test, the series is: absolutely convergent when \(|x-1|<1 \implies -1 < x-1<1 \implies 0<x<2\).

At the endpoints, we have

\(\displaystyle \sum_{k=0}^\infty \frac{(0-1)^k}{\sqrt{n+1}}=\sum_{k=0}^\infty \frac{(-1)^k}{\sqrt{k+1}}\) which converges by the Alternating Series Test
\(\displaystyle \sum_{k=0}^\infty \frac{(2-1)^k}{\sqrt{k+1}}=\sum_{k=1}^\infty\frac{1}{\sqrt{k}}\) which is a p-series with \(p=1/2\) and therefore divergent.

Therefore, the series is convergent on the interval \([0,2)\).

32.2 Learning Objectives

Use manipulation of power series to determine the radius of convergence of a given series.

32.3 Motivation

Given a function, recall that we’re interested in knowing the radius of convergence of its Taylor series; that is, we want to know the interval on which the Taylor series does s good job of approximating the function. We’ve seen the Taylor series of quite a few functions so far; here are a few of our favourites:

\(\displaystyle \sin(x)=\sum_{k=0}^\infty \frac{(-1)^kx^{2k+1}}{(2k+1)!}\) has \(R=\infty\) and therefore converges on \((-\infty, \infty)\)
\(\displaystyle \cos(x)=\sum_{k=0}^\infty \frac{(-1)^kx^{2k}}{(2k)!}\) has \(R=\infty\) and therefore converges on \((-\infty, \infty)\)
\(\displaystyle e^x=\sum_{k=0}^\infty \frac{x^k}{k!}\) has \(R=\infty\) and therefore converges on \((-\infty, \infty)\)
\(\displaystyle \frac{1}{1-x}=\sum_{k=0}^\infty x^k\) has \(R=1\) and therefore converges on \((-1, 1)\)

A new friend that we’ll add to the list is the binomial series:

\[\begin{align*} (1+x)^m &= 1+ mx +m(m-1)\frac{x^2}{2!}+m(m-1)(m-2)\frac{x^3}{3!}+\cdots +m(m-1)(m-2)\cdots(m-n+1)\frac{x^n}{n!}+\cdots \\ &= \sum_{k=0}^\infty {m \choose k} x^k \end{align*}\]

Now, what if we’re given a brand new function and are asked to find the radius of convergence of its Taylor series? The good news is that we can usually avoid a lot of our previous work going to waste; rather than calculating its Taylor polynomial and then testing its convergence, we can use the information from the functions above and some manipulations including differentiation, integration, addition, and subtraction to save time.

Let’s state our main result:

Theorem 32.1 If the series \(\sum c_k(x-x_0)^k\) has radius of convergence \(R\), then we can

differentiate it term-by-term;
integrate it term-by-term;
multiply it through by a constant term-by-term;
add it term-by-term to another series of radius of convergence \(\geq R\)

and the result will also have radius of convergence \(R\).

Note that although the theorem states that the radius of convergence remains the same, the interval of convergence may change.

32.4 Examples

Exercise 32.2 Find the Maclaurin series for \(f(x)=\dfrac{1}{(1-x)^2}\) and determine its interval and radius of convergence.

Solution. We notice that \(f(x)\) is the derivative of the geometric series \(\dfrac{1}{x-1}\), which we know to have radius of convergence \(R=1\).

Differentiating term-by-term for \(|x|<1\), we get

\[\begin{align*} \frac{d}{dx}\left (\frac{1}{1-x} \right) &= \frac{d}{dx} (1+x+x^2+x^3+x^4+\cdots)\\ \frac{1}{(1-x)^2}&=0+1+2x+3x^2+4x^3+\cdots\\ &=\sum_{k=1}^\infty kx^{k-1} \end{align*}\]

Since we took derivatives, our theorem about manipulating power series tells us that the radius of convergence is still \(R=1\).

We check the endpoints:

At \(x=-1\) we have \(\sum_{k=1}^\infty k(-1)^{k-1}\) which diverges
At \(x=1\) we have \(\sum_{k=1}^\infty k 1^{k-1} =\sum_{k=1}^\infty k\) which diverges

Therefore the interval of convergence is \((-1,1)\).

Exercise 32.3 Find the Maclaurin series for \(f(x)=\ln(1-x)\) and determine its interval and radius of convergence.

Solution. Note that \(f(x)\) can be obtained by integrating the geometric series \(\dfrac{1}{1-x}\) which has radius of convergence \(R=1\) and then multiplying by the constant \(-1\)

We have

\[\begin{align*} \ln(1-x) &= -\int \frac{1}{1-x}dx\\ &= -\int \sum_{k=0}^\infty x^k dx\\ &= -\sum_{k=0}^\infty \frac{x^{k+1}}{k+1} +C \\ &= \sum_{k=0}^\infty \frac{-x^{k+1}}{k+1} +C \\ \end{align*}\]

To find the value of \(C\), we can plug in \(x=0\) to get \(C=\ln(1)=0\).

Therefore, we have \(\ln(1-x)=\sum_{k=0}^\infty \frac{-x^{k+1}}{k+1}\).

We check the endpoints:

At \(x=-1\) we have \(\sum_{k=1}^\infty \frac{-(-1)^{k+1}}{k+1}\) which converges
At \(x=1\) we have \(\sum_{k=1}^\infty \frac{-1^{k+1}}{k+1}\) which diverges

Therefore the interval of convergence is \([-1,1)\).

Exercise 32.4 Find the Maclaurin series for \(f(x)=e^x+ln(1-x)\) and determine its radius of convergence.

Solution. Using the Maclaurin series for \(e^x\) and the series for \(\ln(1-x)\) which we found in the previous example, we have

\[\begin{align*} e^x+\ln(1-x) &= \sum_{k=0}^\infty \frac{x^k}{k!} - \sum_{k=0}^\infty \frac{x^{k+1}}{k+1}\\ &= \sum_{k=0}^\infty \frac{x^k}{k!} - \sum_{k=1}^\infty \frac{x^k}{k} \quad\mbox{re-indexing the second sum}\\ &= 1+ \sum_{k=1}^\infty \frac{x^k}{k!} - \sum_{k=1}^\infty \frac{x^k}{k} \quad\mbox{re-indexing the first sum}\\ &= 1+\sum_{k=1}^\infty \left ( \frac{1}k! -\frac{1}{k} \right) x^k \end{align*}\]

Note that the radius of convergence of \(e^x\) is \(R_1=\infty\) and the radius of convergence of \(\ln(1-x)\) is \(R_2=1\); we choose the smaller radius \(R=1\).